Bài 5 :

a) \(\dfrac{y}{4}=\dfrac{9}{y}\)

\(\Rightarrow y^2=36\left(y\ne0\right)\)

\(\Rightarrow y=\pm6\)

b) \(\dfrac{y+7}{20}=\dfrac{5}{y+7}\left(y\ne-7\right)\)

\(\Rightarrow\left(y+7\right)^2=100=10^2\)

\(\Rightarrow\left[{}\begin{matrix}y+7=10\\y+7=-10\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}y=3\\y=-17\end{matrix}\right.\)

c) \(\dfrac{4-5y}{3}=\dfrac{y+2}{5}\)

\(\Rightarrow5\left(4-5y\right)=3\left(y+2\right)\)

\(\Rightarrow20-25y=3y+6\)

\(\Rightarrow28y=14\)

\(\Rightarrow y=\dfrac{14}{28}=\dfrac{1}{2}\)

Bài 4 :

\(\dfrac{a}{5}=\dfrac{b}{7}=\dfrac{c}{10}\)

\(\Rightarrow\dfrac{2a}{10}=\dfrac{3b}{21}=\dfrac{4c}{40}=\dfrac{2a+3b-4c}{10+21-40}=\dfrac{81}{-9}=-9\)

\(\Rightarrow\left\{{}\begin{matrix}a=-9.5=-45\\b=-9.7=-63\\c=-9.10=-90\end{matrix}\right.\)

Đính chính Bài 4 :

\(...\dfrac{2a+3b-4c}{10+21-40}=\dfrac{-81}{-9}=9\)

\(\Rightarrow\left\{{}\begin{matrix}a=9.4=36\\b=9.7=63\\c=9.10=90\end{matrix}\right.\)

a: \(\left(x,y\right)\in\left\{\left(-9;1\right);\left(-1;9\right);\left(-3;3\right)\right\}\)

b: \(\left(x,y\right)\in\left\{\left(1;7\right);\left(-7;-1\right)\right\}\)

c: \(\left(x,y\right)\in\left\{\left(11;-1\right);\left(-11;1\right)\right\}\)

a: 

b: 

c: 

a: \(\left(x,y\right)\in\left\{\left(-9;1\right);\left(-1;9\right);\left(-3;3\right)\right\}\)

b: \(\left(x,y\right)\in\left\{\left(1;7\right);\left(-7;-1\right)\right\}\)

c: \(\left(x,y\right)\in\left\{\left(11;-1\right);\left(-1;11\right)\right\}\)

Ta có :

- x/3 = y/7 suy ra : x/6 = y/14

- y/2 = z/5 suy ra : y/14 = z/35

Và ................................

Kết quả là : x = 24 ; z = 140

ai tk mk mk tk lại

Ta có:

- x/3 = y/7 suy ra: x/6 = y/14

- y/2 = z/5 suy ra: y/14 = z/35

Và.......................................................

Nói chung kết quả: x=24

                             y=56

                             z=140

Theo đề bài ta có\(\frac{3+x}{7+y}=\frac{3}{7}\)

<=>\(\frac{3.a}{7.a}=\frac{3+x}{7+y}=\frac{3}{7}\)

x=20:(3+7).3=6

y=20:(3+7).7=14

vậy x= 6 ; y = 14